We are using the formula P=%(W) (Part = Percent x Whole)

Write y=5/8x +6 in standard form using intergers

Art [367]

Answer:

5x - 8y = -48

Step-by-step explanation:

Move 5/8x to the y side to get (-5/8)x + y = 6. Multiply both sides by -8 to get 5x - 8y = -48.

8 0

2 years ago

Five college students' scores on a test are 25, 25, 30, 35, and 40. what is the mean for this set of scores?

klio [65]

Answer:

Mean= 31

Step-by-step explanation:

To find the mean you need to add all the number together and then divide by how many ever numbers there are.

In this case

25 + 25 + 30 + 35 + 40=155

There are 5 numbers so you would divide by 5

155/5

=31

5 0

1 year ago

Lynnette can wash 959595 cars in 5 days.<br> How many cars can Lynnette wash in 11 days?

aleksley [76]

Answer:

Lynnette can wash 2111109 in 11 days

Step-by-step explanation:

Have a good day :)

6 0

3 years ago

[2+(6times8)]-1 plz help

bezimeni [28]

[2+(6times8)]-1

= [2+(6 * 8)] -1

= [2+(48)] -1

= 50 -1

= 49

8 0

3 years ago

Read 2 more answers

Ten years ago 53% of American families owned stocks or stock funds. Sample data collected by the Investment Company Institute in

Alborosie

Answer:

a) Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

b) $z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

$p_v =P(Z$

c) So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

Step-by-step explanation:

Data given and notation

n=300 represent the random sample taken

$\hat p=0.46$ estimated proportion of American families owning stocks or stock funds

$p_o=0.53$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

Part a

We need to conduct a hypothesis in order to test the claim that proportion is less than 0.53 or 53%.:

Null hypothesis: $p\geq 0.53$

Alternative hypothesis: $p < 0.53$

Part b

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.46 -0.53}{\sqrt{\frac{0.53(1-0.53)}{300}}}=-2.429$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a left tailed test the p value would be:

$p_v =P(Z$

Part c

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of American families owning stocks or stock funds is significantly less than 0.53 .

7 0

3 years ago